Distinct permutations of the string | Set 2

Print all distinct permutation of a string having duplicates.

Examples:

Input : ABCA
Output : AABC AACB ABAC ABCA ACBA 
         ACAB BAAC BACA BCAA CABA 
         CAAB CBAA

An algorithm to print all distinct permutations has already been discussed here. Here we’ll discuss one more approach to do the same. Recall first how we print permutations without any duplicates in the input string. It is given here. Let’s now take the case of the string “ABAC”. While generating permutations, let’s say we are at index = 0, swap it with all elements after it. When we reach at i=2, we see that in the string s[index…i-1], there was an index which is equal to s[i]. Thus, swapping it will produce repeated permutations. Thus, we don’t swap it. The below explains it better.



Illustration : Let us understand with below example.

i = 0 1 2 3
    A B A C
index = 0, s[0] = A
Start swapping s[index] with s[i] following it:
i = index + 1 = 1 

Since s[index] != s[i], swap and recur.

i = 2, s[index] == s[i], don't swap

i = 3,  s[index] != s[i], swap and recur. 

Below code does the same.

C++

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// C++ program to distinct permutations of the string
#include <bits/stdc++.h>
using namespace std;
  
// Returns true if str[curr] does not matches with any of the
// characters after str[start]
bool shouldSwap(char str[], int start, int curr)
{
    for (int i = start; i < curr; i++) 
        if (str[i] == str[curr])
            return 0;
    return 1;
}
  
// Prints all distinct permutations in str[0..n-1]
void findPermutations(char str[], int index, int n)
{
    if (index >= n) {
        cout << str << endl;
        return;
    }
  
    for (int i = index; i < n; i++) {
  
        // Proceed further for str[i] only if it 
        // doesn't match with any of the characters
        // after str[index]
        bool check = shouldSwap(str, index, i);
        if (check) {
            swap(str[index], str[i]);
            findPermutations(str, index + 1, n);
            swap(str[index], str[i]);
        }
    }
}
  
// Driver code
int main()
{
    char str[] = "ABCA";
    int n = strlen(str);
    findPermutations(str, 0, n);
    return 0;
}

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Java

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// Java program to distinct permutations of the string
public class GFG {
  
// Returns true if str[curr] does not matches with any of the 
// characters after str[start] 
    static boolean shouldSwap(char str[], int start, int curr) {
        for (int i = start; i < curr; i++) {
            if (str[i] == str[curr]) {
                return false;
            }
        }
        return true;
    }
  
// Prints all distinct permutations in str[0..n-1] 
    static void findPermutations(char str[], int index, int n) {
        if (index >= n) {
            System.out.println(str);
            return;
        }
  
        for (int i = index; i < n; i++) {
  
            // Proceed further for str[i] only if it 
            // doesn't match with any of the characters 
            // after str[index] 
            boolean check = shouldSwap(str, index, i);
            if (check) {
                swap(str, index, i);
                findPermutations(str, index + 1, n);
                swap(str, index, i);
            }
        }
    }
  
    static void swap(char[] str, int i, int j) {
        char c = str[i];
        str[i] = str[j];
        str[j] = c;
    }
  
    // Driver code
    public static void main(String[] args) {
  
        char str[] = {'A', 'B', 'C', 'A'};
        int n = str.length;
        findPermutations(str, 0, n);
    }
  
}

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Python3

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# Python3 program to distinct
# permutations of the string 
  
# Returns true if str[curr] does not 
# matches with any of the characters
# after str[start] 
def shouldSwap(string, start, curr): 
  
    for i in range(start, curr): 
        if string[i] == string[curr]: 
            return 0
    return 1
  
# Prints all distinct permutations
# in str[0..n-1] 
def findPermutations(string, index, n): 
  
    if index >= n: 
        print(''.join(string)) 
        return
  
    for i in range(index, n): 
  
        # Proceed further for str[i] only 
        # if it doesn't match with any of 
        # the characters after str[index] 
        check = shouldSwap(string, index, i) 
        if check: 
            string[index], string[i] = string[i], string[index] 
            findPermutations(string, index + 1, n) 
            string[index], string[i] = string[i], string[index] 
  
# Driver code 
if __name__ == "__main__":
  
    string = list("ABCA"
    n = len(string) 
    findPermutations(string, 0, n) 
      
# This code is contributed by Rituraj Jain

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C#

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// C# program to distinct permutations
// of the string
using System;
  
class GFG
  
// Returns true if str[curr] does 
// not matches with any of the 
// characters after str[start] 
static bool shouldSwap(char []str, 
                       int start, int curr) 
    for (int i = start; i < curr; i++) 
    
        if (str[i] == str[curr]) 
        
            return false
        
    
    return true
  
// Prints all distinct permutations 
// in str[0..n-1] 
static void findPermutations(char []str, 
                             int index, int n) 
    if (index >= n) 
    
        Console.WriteLine(str); 
        return
    
  
    for (int i = index; i < n; i++) 
    
  
        // Proceed further for str[i] only 
        // if it doesn't match with any of 
        // the characters after str[index] 
        bool check = shouldSwap(str, index, i); 
        if (check) 
        
            swap(str, index, i); 
            findPermutations(str, index + 1, n); 
            swap(str, index, i); 
        
    
  
static void swap(char[] str, int i, int j)
    char c = str[i]; 
    str[i] = str[j]; 
    str[j] = c; 
  
// Driver code 
public static void Main() 
    char []str = {'A', 'B', 'C', 'A'}; 
    int n = str.Length; 
    findPermutations(str, 0, n); 
  
// This code is contributed 
// by 29AjayKumar

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Output:

ABCA
ABAC
ACBA
ACAB
AACB
AABC
BACA
BAAC
BCAA
CBAA
CABA
CAAB

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